Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 110;
int a[N][N], r[N], c[N];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        scanf("%d", &a[i][j]);
      }
    }
    int ans(-1), r0;
    for (int k = 0; k <= a[0][0]; ++k) {
      r[0] = k;
      bool f = true;
      for (int j = 0; j < m; ++j) {
        c[j] = a[0][j] - r[0];
        if (c[j] < 0) f = false;
      }
      if (!f) continue;
      for (int i = 1; i < n; ++i) {
        r[i] = a[i][0] - c[0];
        for (int j = 0; j < m; ++j) {
          if (a[i][j] - c[j] != a[i][0] - c[0]) f = false;
        }
        if (r[i] < 0) f = false;
        if (!f) break;
      }
      if (!f) continue;
      int tmp(0);
      for (int i = 0; i < n; ++i) tmp += r[i];
      for (int j = 0; j < m; ++j) tmp += c[j];
      if ((ans == -1 || ans > tmp)) ans = tmp, r0 = k;
    }
    if (ans != -1) {
      printf("%d\n", ans);
      r[0] = r0;
      for (int j = 0; j < m; ++j)
        if (a[0][j] - r[0]) {
          c[j] = a[0][j] - r[0];
          for (int k = 0; k < c[j]; ++k) printf("col %d\n", j + 1);
        }
      for (int i = 0; i < n; ++i)
        for (int k = 0; k < a[i][0] - c[0]; ++k) printf("row %d\n", i + 1);
    } else {
      puts("-1");
    }
  }
  return 0;
}